On Pillai's Problem involving Lucas sequences of the second kind

In this paper we consider the Diophantine equation $ V_n - b^m = c $ for given integers $ b,c $ with $ b \geq 2 $, whereas $ V_n $ varies among Lucas-Lehmer sequences of the second kind. We prove under some technical conditions that if the considered equation has at least three solutions $ (n,m) $, then there is an upper bound on the size of the solutions as well as on the size of the coefficients in the characteristic polynomial of $ V_n $.


Introduction
In recent time many authors considered Pillai-type problems involving linear recurrence sequences.For an overview we refer to [5].Let us note that these problems are inspired by a result due to S. S. Pillai [9,10] who proved that for given, coprime integers a and b there exists a constant c 0 (a, b), depending on a and b, such that for any c > c 0 (a, b) the equation (1) a n − b m = c has at most one solution (n, m) ∈ Z 2 >0 .Replacing the powers a n and b m by other linear recurrence sequences seems to be a challenging task which was supposedly picked up first in [4], where it was shown that F n − 2 m = c has at most one solution (n, m) ∈ Z 2 >0 provided that c / ∈ {0, 1, −1, −3, 5, −11, −30, 85}.
More generally, Chim, Pink and Ziegler [3] proved that for two fixed linear recurrence sequences (U n ) n∈N , (V n ) n∈N (with some restrictions) the equation has at most one solution (n, m) ∈ Z 2 >0 for all c ∈ Z, except if c is in a finite and effectively computable set C ⊂ Z that depends on (U n ) n∈N and (V n ) n∈N .
In more recent years several attempts were made to obtain uniform results, i.e. to allow to vary the recurrence sequences (U n ) n∈N and (V n ) n∈N in the result of Chim, Pink and Ziegler [3].In particular, Batte et.al. [1] showed that for all pairs (p, c) ∈ P × Z with p a prime the Diophantine equation has at most four solutions (n, m) ∈ N 2 with n ≥ 2. This result was generalized by Heintze et.al. [5].They proved (under some technical restrictions) that for a given linear recurrence sequence (U n ) n∈N there exist an effectively computable bound B ≥ 2 such that for an integer b > B the Diophantine equation (2) U n − b m = c has at most two solutions (n, m) ∈ N 2 with n ≥ N 0 .Here N 0 is an effective computable constant depending only on (U n ) n∈N .
In this paper we want to fix b in (2) and let (U n ) n∈N vary over a given family of recurrence sequences.In particular we consider the case where (U n ) n∈N varies over the family of Lucas-Lehmer sequences of the second kind.

Notations and statement of the main results
In this paper we consider Lucas-Lehmer sequences of the second kind, that is we consider binary recurrence sequences of the form where α and β are the roots of the quadratic polynomial with A 2 = 4B and gcd(A, B) = 1.In the following we will assume that V n is non-degenerate.That is we assume A 2 − 4B > 0 and A = 0, since this implies that α and β are distinct real numbers with |α| = |β|.We will also assume that A > 0, which results in V n > 0 for all n ∈ N. Then we consider the Diophantine equation Theorem 1.Let 0 < ε < 1 be a fixed real number and assume that |B| < A 2−ε as well as that the polynomial X 2 − AX + B is irreducible.Furthermore, assume that b ≥ 2 if c ≥ 0 and b ≥ 3 if c < 0. Assume that Equation (3) has three solutions (n 1 , m 1 ), (n 2 , m 2 ), (n 3 , m 3 ) ∈ N 2 with n 1 > n 2 > n 3 ≥ N 0 (ε) for the bound N 0 (ε) = 3 2ε .Then there exists an effectively computable constant C 1 = C 1 (ε, b), depending only on ε and b, such that n 1 < C 1 or A < 32 1/ε .In particular, we can choose C 1 = 4.83 • 10 32 (log b) 4  ε 2 log 5.56 Let us note that the bound N 0 = N 0 (ε) in Theorem 1 ensures that V n is strictly increasing for n ≥ N 0 (see Lemma 9 below).Let us also mention that it is essential to exclude the case that V n2 = V n3 .
Although we can bound n 1 in terms of b and ε our method does not provide upper bounds for A and |B|.However, in the case that we are more restrictive in the possible choice of B we can prove also upper bounds for A and |B|.A straight forward application of our bounds yields: Another consequence of Theorem 2 together with the results of Chim et.al. [3] is that there exist only finitely many Diophantine equations of the form of (3) that admit more than two solutions, provided that |B| < κA.The following corollary gives a precise statement.

Corollary 4.
Let κ > 0 be a fixed real number and b ≥ 2 a fixed integer.Then there exist at most finitely many, effectively computable 9-tuples (A, Remark 1.Let us note that for an application of the results of Chim et.al. [3] we have to ensure that α and b are multiplicatively independent.However, α and b are multiplicatively dependent if and only if there exist integers x, y not both zero such that α x = b y .But α x cannot be a rational number unless x = 0 or α = √ D for some positive integer D. But α = √ D would imply A = 0 which we have excluded.Therefore we can apply their results in our situation and obtain the statement of Corollary 4, provided that we have found an upper bound for A. Let us give a quick outline for the rest of the paper.In the next section we establish several lemmas concerning properties of Lucas sequences V n under the restrictions that |B| < A 2−ε and |B| < κA, respectively, that we will frequently use throughout the paper.The main tool for the proofs of the main theorems are lower bounds for linear forms in logarithms of algebraic numbers.In Section 6 we establish bounds for n 1 , which still depend on log α, following the usual approach (cf.[3]).In Section 7, under the assumption that three solution exist, we obtain a system of inequalities involving linear forms in logarithms which contain log α.Combinig these inequalities we obtain a linear form in logarithms which does no longer contain log α.Thus we obtain that Theorems 1 and 2 hold or one of the following two cases occurs: • That each of these cases implies Theorems 1 and 2 is shown in the subsequent Sections 8 and 9.

Auxiliary results on Lucas sequences
Let α and β be the roots of X 2 − AX + B. By our assumptions α and β are distinct real numbers / ∈ Q.Throughout the rest of the paper we will assume that |α| > |β|, which we can do since by our assumptions we have A = 0. Therefore we obtain First note that α > 1 can be bounded in terms of A and ε, respectively in terms of A and κ.

Lemma 5. Assuming that |B|
At some point in our proofs of Theorems 1 and 2 we will use that β cannot be to close to 1.In particular, the following lemma will be needed.Lemma 6. Assume that A ≥ 4. Then we have Proof.First, let us note that the function f is strictly monotonically increasing for all x < A 2 /4.Moreover, we have f (A − 1) = 1 and f (−A − 1) = −1.By our assumption that X 2 − AX + B is irreducible, a choice for B such that |β| = 1 is not admissible.Therefore we have Next, we show that under our assumptions |β| is not to large.
Proof.Let us assume that A ε ≥ 8. First we note that , which is defined for all x with 0 < 4x ≤ A 2 .Then we have Note that it is immediate that f ′ (x) < 0 for x > 0. Thus we deduce since we assume that A ε ≥ 8.
Proof.We assume that A ≥ 4κ.Moreover, we have Let us now consider the function This function is strictly increasing with x since obviously the numerator is strictly increasing with x while the denominator is strictly decreasing with x.Therefore we obtain Now let us take a look at the recurrence sequence V n .
Since the rational function f (x) = ax+b cx+d is strictly increasing if ad − bc > 0, strictly decreasing if ad − bc < 0, and constant if ad − bc = 0, we obtain that for the Lucas sequence is strictly increasing.
In view of this remark and in view of the proofs of the following lemmata, we will assume for the rest of the paper that one of the following two assumptions holds: Remark 3. Let us note that the bound A ≥ κ 2 and A ≥ 1024 in Assumption A2 results in the useful fact that Assumption A2 implies Assumption A1 with ε = 12 , but with N 0 = 1 instead of 3 2ε .The assumption A ≥ 16κ + 12 is mainly used in the proofs of the Lemmata 12 and 13.
In view of the assumptions stated above an important consequence of Lemmata 9 and 10 is the following: Lemma 12. Let n ≥ N 0 , then we have 5  4 α n > V n > 3 4 α n .Proof.Assume that A1 holds.Then, by Lemma 7, we have If A2 holds, by Lemma 8, we have Therefore in any case we get Another lemma that will be used frequently is the following: Proof.Assuming the existence of two different solutions implies by an application of Lemma 12 the inequality 5 4 which proves the first inequality.
For the second inequality we apply again Lemma 12 to obtain

Since we assume in any case that α >
which yields the second inequality.Proof.Assume that (3) has two solutions (n 1 , m 1 ), (n 2 , m 2 ) with n 1 > n 2 ≥ 1.Then by Carmichael's primitive divisor theorem (Lemma 14) we deduce that n 1 = 3 and n 2 = 1.Since V 1 = A and V 3 = A 3 − 3AB we obtain the system of equations That is b | 3B.Since we assume that gcd(A, B) = 1, we deduce b | 3, i.e. b = 3.We also conclude that 3 ∤ B which yields, by considering 3-adic valuations, that Note that we also assume A 2 − 4B > 0 which implies that holds.But this is only possible for m 2 = 1 and we conclude that is not irreducible, this is not an admissible case.Therefore there exists at most one solution to (3).

Lower bounds for linear forms in logarithms
The main tool in proving our main theorems are lower bounds for linear forms of logarithms of algebraic numbers.In particular, we will use Matveev's lower bound proven in [7].Therefore let η = 0 be an algebraic number of degree δ and let be the minimal polynomial of η.Then the absolute logarithmic Weil height is defined by With this notation the following result due to Matveev [7] holds: Lemma 16 (Theorem 2.2 with r = 1 in [7]).Denote by η 1 , . . ., η N algebraic numbers, neither 0 nor 1, by log η 1 , . . ., log η N determinations of their logarithms, by D the degree over Q of the number field and set Assume that In our applications we will be in the situation N ∈ {2, 3} and Remark 4. Let us note that the form of E is essential in our proof to obtain an absolute bound for n 1 in Theorem 1.Let us also note that in the case of N = 2 one could use the results of Laurent [6] to obtain numerically better values but with an log(E) 2 term instead.This would lead to numerically smaller upper bounds for concrete applications of our theorems.However, we refrain from the application of these results to keep our long and technical proof more concise.
In order to apply Matveev's lower bounds, we provide some height computations.First, we note the following well known properties of the absolute logarithmic height for any η, γ ∈ Q and l ∈ Q (see for example [12,Chapter 3] for a detailed reference): Moreover, note that for a positive integer b we have h(b) = log b and This together with the above mentioned properties yields the following lemma: Lemma 18.Under the assumptions A1 or A2 and using the above notations from Lemma 16 the following inequalites hold for any t ∈ Z >0 : One other important aspect in applying Matveev's result (Lemma 16) is that the linear form Λ does not vanish.We will resolve this issue with the following lemma: Lemma 19.Assume that the Diophantine equation (3) has three solutions (n 1 , m 1 ), (n 2 , m 2 ), (n 3 , m 3 ) ∈ N 2 with n 1 > n 2 > n 3 > 0. Then we have This implies α n1 − α n2 = b m1 − b m2 which results in view of (7) in But then β = 0 or β is a root of unity.Both cases contradict our assumption that Finally, we want to record three further elementary lemmata that will be helpful.The first lemma is a standard fact from real analysis.Proof.A direct application of Taylor's theorem with a Cauchy and Lagrange remainder, respectively.
Next, we want to state another estimate from real analysis: Lemma 21.Let x, n ∈ R such that |2nx| < 0.5 and n ≥ 1.Then we have Proof.Since e y is a convex function, we have for 0 ≤ y < 0.5 that e y ≤ 1 + y e 0.5 − 1 and for −0.5 < y ≤ 0 we obtain That is we have |1 − e y | ≤ 1.3|y| for |y| < 0.5.Note that by our assumptions we have |x| < 0.5 and therefore we obtain by an application of Lemma 20 The third lemma is due to Pethő and de Weger [8].
A proof of this lemma can be found in [11, Appendix B].

A lower bound for |c| in terms of n 1 and α
The purpose of this section is to prove a lower bound for |c|.In particular we prove the following proposition: Proof.Let us assume that, in contrary to the content of the proposition, We consider equation The goal is to apply Matveev's theorem (Lemma 16) with N = 2.Note that with η 1 = b and η 2 = α we choose A 1 = 2 log b and A 2 = 2 log α, in view of Lemma 18, and obtain Due to Lemma 13 we have Therefore we obtain by Corollary 17 that 2.68 which implies the content of the proposition.

Bounds for n 1 in terms of log α
In this section we will assume that Assumption A1 or A2 holds.However, in the proofs we will mainly consider the case that Assumption A1 holds.Note that this is not a real restriction since Assumption A2 implies Assumption A1 with ε = 1 2 and N 0 = 1 instead of N 0 = 3 2ε (cf.Remark 3).Also assume that Diophantine equation (3) has three solutions (n 1 , m 1 ), (n 2 , m 2 ), (n 3 , m 3 ) with n 1 > n 2 > n 3 ≥ N 0 .In this section we follow the approach of Chim et.al. [3] and prove upper bounds for n 1 in terms of α.To obtain explicit bounds and to keep track of the dependence on log b and log α of the bounds we repeat their proof.This section also delivers the set up for the later sections which provide proofs of our main theorems.Moreover, note that the assumption that three solutions exist, simplifies the proof of Chim et.al. [3].
The main result of this section is the following statement: Proposition 24.Assume that Assumption A1 or A2 holds and that Diophantine equation (3) has three solutions (n 1 , m 1 ), (n 2 , m 2 ), (n 3 , m 3 ) with n 1 > n 2 > n 3 ≥ N 0 .Then we have where we choose ε = 1/2 in case that Assumption A2 holds.
Since we assume the existence of two solutions, we have and therefore obtain Let us write γ := min{α, α/|β|}.Note that we have γ > 1 2 A ε > 1 4 α ε , by Lemma 7 and Lemma 5.With this notation we get the inequality Note that, depending on whether |β| > 1 or |β| ≤ 1, we have Therefore, using Lemma 13, we obtain First, let us assume that the maximum in ( 8) is γ −n1 .Under our assumptions we have A ε ≥ 32 and n 1 ≥ 3, i.e. 7γ −n1 < 1  2 .Thus taking logarithms and applying We apply Matveev's theorem (Lemma 16) with N = 2.Note that with η 1 = b and η 2 = α we choose A 1 = 2 log b and A 2 = 2 log α, in view of Lemma 18, to obtain Note that, due to Lemma 13, Therefore For the rest of the proof of Proposition 24 we will assume that (9) holds.Since we assume that a third solution exists, the statement of Lemma 13 also holds for m 2 and n 2 instead of m 1 and n 1 .In particular we have Let us rewrite Equation (7) again to obtain the inequality As previously noted we have 4γ −n1 < 1 2 and therefore we obtain We aim to apply Matveev's theorem to Λ ′ with η Note that due to Lemma 18 and the properties of heights we obtain Thus we obtain E ≤ 2n 1 as before, and from Matveev's theorem which yields the content of Proposition 24.
Let us reconsider Inequality (8) with n 1 , m 1 , n 2 , m 2 replaced by n 2 , m 2 , n 3 , m 3 , respectively.Then we obtain Note that with the assumption that three solutions exist we can apply Lemma 13 to b m 2 α n 2 but not to b m 3 α n 3 .Let us assume for the next paragraphs that (13) Then, by applying Lemma 20 to ( 8) and ( 12), we obtain the system of inequalities We eliminate the term log α from these inequalities by considering Λ 0 = n 2 Λ 1 −n 1 Λ 2 and obtain the inequality Let us write M for the maximum on the right hand side of ( 14).If n 2 m 1 − n 1 m 2 = 0, we obtain the inequality log b ≤ M .Since we will study the case that n 2 m 1 − n 1 m 2 = 0 in Section 8, we will assume for the rest of this section that n 2 m 1 − n 1 m 2 = 0, i.e. we have log b ≤ M .Therefore we have to consider five different cases.In each case we want to find an upper bound for log α if possible: The case M = 12n To obtain from this inequality a bound for log α is not straight forward and we will deal with this case in Section 9.The case M = 28n 2 α n2−n1 : By a similar computation as in the first case, we obtain in this case the inequality log α < log(40.4n 2 ) < log(40.4n 1 ).
In the case that (13) does not hold, i.e. that M 0 ≥ 1/2, we obtain by similar computations in each of the five possibilities the following inequalities: The case M 0 = 7α n2−n1 : log α ≤ log 14; The case M 0 = 4γ −n1 : log α ≤ log 64 ε .Let us recap what we have proven so far in the following lemma: Lemma 25.Assume that Assumption A1 or A2 holds and assume that Diophantine equation (3) has three solutions (n 1 , m 1 ), (n 2 , m 2 ), (n 3 , m 3 ) with n 1 > n 2 > n 3 ≥ N 0 .Then one of the following three possibilities holds: Since we will deal with the first and second possiblity in the next sections, we close this section by proving that the last possibility implies Theorems 1 and 2. Therefore let us plug in the upper bound for log α into Inequality (11)  .
This proves Theorem 2 in the current case.

The case
We distinguish between the cases c ≥ 0 and c < 0.
8.1.The case c ≥ 0. In this case we have holds under our assumptions, we may apply Lemma 20 to get the two inequalities Multiplying the first inequality by n 2 and the second one by n 1 as well as forming the difference afterwards yields Let us note that (a + b) 2 ≤ 4 max{|a| 2 , |b| 2 }, and therefore we obtain Together with the estimate Let us assume for the moment that the maximum is c 2 α n 2 .Then we obtain α n2 < 12c < 24α n3 which implies α < 24 and thus Theorem 2. Plugging in α < 24 in Proposition 24 yields the content of Theorem 1 in this case.
Therefore we assume now c < 12|β| n2 .By Proposition 23 we obtain Note that, due to our assumptions, we have the bound As c < 2α n3 (see Inequality (15)) we obtain which yields α < 20κ.If c < β n2 , then (16) gives us and hence α < 8κ.Thus we may now assume |c − β n2 | < 2(c + |β| n1 )α n2−n1 .Let us note that under the assumption α > 2(c + max{1, |β|} n1 ) we can deduce and otherwise we would get the constant C 2 in Theorem 2 (cf. the calculations below).Then, by Lemma 20, we get Recalling from the beginning of Section 7 the bound we can again eliminate the term log α from these inequalities by considering the form Λ ′ 0 = n 3 Λ 1 − n 1 Λ 3 and obtain the inequality If From ( 21) we deduce that one of the two factors of the left hand side is smaller than 2α . By thinking of constant C 2 in Theorem 2, we may assume α > 64(c + |β| n1 ).So we have 2α This implies |β| n3 < 1 4 and further Therefore we have c = 0.But Lemma 15 states that there cannot be three solutions for c = 0. Now we may assume Note that, in particular, we have β n2−n3 > 0, and the above inequality implies Using the binomial series expansion for (1 ± x) r with exponent r Assuming α > 64n 2 2 (c + |β| n1 ), we obtain by an application of Lemma 21 that This together with our assumption ).Thus we may assume c = 1 provided that α is large enough.But this also implies ). Therefore we may assume β n3 = |β| n3 is positive.Since for any real numbers x > 0 and n ≥ 1 we have |1 − x| ≤ |1 − x n |, we obtain from Lemma 6 together with Lemma 5 2 4α + 5 Hence we get α < 9(c + |β| n1 ) in this case.
Let us summarize what we have proven so far: In the case c ≥ 0 under Assumption A2 we cannot have three solutions to (3) for α ≥ 433n 2 2 (c + |β| n1 ).So if there exist three solutions, then we have 8.2.The case c < 0. The case c < 0 can be treated with similar arguments.Therefore we will only point out the differences.
We start with the observation and write again b holds under our assumptions if in addition m 2 − m 3 ≥ 2. The case m 2 − m 3 = 1 is included in the next section.Thus we get again the inequality chain (16) and furthermore the bound which implies m 2 − m 3 ≤ 3.This will be handled in Section 9. Therefore we may now again assume |c| < 12|β| n2 .In the same way as in the case c ≥ 0 we obtain again the upper bound (18) proving Theorem 1 also in the case c < 0.Moreover, we get under Assumption A2 the same upper bound (19) for n 1 .In particular, the quantities |c|, |β| n1 , |β| n2 are bounded by absolute, effectively computable constants.
Apart from the special case which, by Lemma 13, yields 1 15 In view of Theorems 1 and 2 and Lemma 26 we may assume that Assumption A1 or A2 holds and that m 2 − m 3 < 1.45 log(43.3n 1 ).
First we reconsider inequality (8) and note that 7 max {α n2−n1 , γ −n1 } ≥ 1 2 implies either α ≤ 14 or A ε ≤ 28.In the first cases we have an upper bound for α and by Proposition 24 also an upper bound for n 1 ; the second case contradicts Assumption A1 and A2 respectively.Thus Theorems 1 and 2 are shown in those situations.Now we may apply Lemma 20 and obtain, as in Section 6, the inequality Next, let us consider the inequality ≤ 18 max{α n3−n2 , γ −n2 }.

( 3 )
V n − b m = c, where b, c ∈ Z with b > 1 are fixed.

Finally, let us
remind Carmichael's theorem [2, Theorem XXIV]: Lemma 14.For any n = 1, 3 there exists a prime 1 p such that p | V n , but p ∤ V m for all m < n, except for the case that n = 6 and (A, B) = (1, −1).This lemma can be used to prove the following result: Lemma 15.Assume that c = 0 and A > 1.Then the Diophantine equation (3) has at most one solution (n, m) with n ≥ 1.
irreducible, α and β are Galois conjugated.Therefore, by applying the non-trivial automorphism of K = Q(α) to the equation β ni = c, we obtain α ni = c since c ∈ Q.But this implies β ni = α ni , hence |α| = |β|, a contradiction to our assumptions.Now, let us assume that